Answer
$0.08 J$ or 0.08 N-m
Work Step by Step
Since, we know $F=k x$
Then, $k=\dfrac{F}{x}=\dfrac{2}{0.02}= 100$ N-m
The rubber band will stretch, that is, the limits for $x$ will be from $0$ to:
$x=\dfrac{4}{100}=0.04 m$
Now, work done will be
$W=\int_0^{0.04} 100 x dx= 100[ \dfrac{x^2}{2}]_0^{0.04}$
or, $W=50[ (0.04)^2-0] =0.08 J$ or 0.08 N-m
